3.1040 \(\int x^5 (a+b x^2)^p \, dx\)

Optimal. Leaf size=72 \[ \frac{a^2 \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac{a \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{\left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)} \]

[Out]

(a^2*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) - (a*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (a + b*x^2)^(3 + p)/(2*b^3
*(3 + p))

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Rubi [A]  time = 0.0431496, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {266, 43} \[ \frac{a^2 \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac{a \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac{\left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*x^2)^p,x]

[Out]

(a^2*(a + b*x^2)^(1 + p))/(2*b^3*(1 + p)) - (a*(a + b*x^2)^(2 + p))/(b^3*(2 + p)) + (a + b*x^2)^(3 + p)/(2*b^3
*(3 + p))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int x^5 \left (a+b x^2\right )^p \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{a^2 (a+b x)^p}{b^2}-\frac{2 a (a+b x)^{1+p}}{b^2}+\frac{(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac{a \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac{\left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}\\ \end{align*}

Mathematica [A]  time = 0.0272372, size = 64, normalized size = 0.89 \[ \frac{\left (a+b x^2\right )^{p+1} \left (2 a^2-2 a b (p+1) x^2+b^2 \left (p^2+3 p+2\right ) x^4\right )}{2 b^3 (p+1) (p+2) (p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*x^2)^p,x]

[Out]

((a + b*x^2)^(1 + p)*(2*a^2 - 2*a*b*(1 + p)*x^2 + b^2*(2 + 3*p + p^2)*x^4))/(2*b^3*(1 + p)*(2 + p)*(3 + p))

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Maple [A]  time = 0.004, size = 80, normalized size = 1.1 \begin{align*}{\frac{ \left ( b{x}^{2}+a \right ) ^{1+p} \left ({b}^{2}{p}^{2}{x}^{4}+3\,{b}^{2}p{x}^{4}+2\,{b}^{2}{x}^{4}-2\,abp{x}^{2}-2\,ab{x}^{2}+2\,{a}^{2} \right ) }{2\,{b}^{3} \left ({p}^{3}+6\,{p}^{2}+11\,p+6 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*x^2+a)^p,x)

[Out]

1/2*(b*x^2+a)^(1+p)*(b^2*p^2*x^4+3*b^2*p*x^4+2*b^2*x^4-2*a*b*p*x^2-2*a*b*x^2+2*a^2)/b^3/(p^3+6*p^2+11*p+6)

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Maxima [A]  time = 2.04432, size = 99, normalized size = 1.38 \begin{align*} \frac{{\left ({\left (p^{2} + 3 \, p + 2\right )} b^{3} x^{6} +{\left (p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + 2 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{p}}{2 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^p,x, algorithm="maxima")

[Out]

1/2*((p^2 + 3*p + 2)*b^3*x^6 + (p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + 2*a^3)*(b*x^2 + a)^p/((p^3 + 6*p^2 + 11*p
+ 6)*b^3)

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Fricas [A]  time = 1.59954, size = 197, normalized size = 2.74 \begin{align*} \frac{{\left ({\left (b^{3} p^{2} + 3 \, b^{3} p + 2 \, b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} +{\left (a b^{2} p^{2} + a b^{2} p\right )} x^{4} + 2 \, a^{3}\right )}{\left (b x^{2} + a\right )}^{p}}{2 \,{\left (b^{3} p^{3} + 6 \, b^{3} p^{2} + 11 \, b^{3} p + 6 \, b^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^p,x, algorithm="fricas")

[Out]

1/2*((b^3*p^2 + 3*b^3*p + 2*b^3)*x^6 - 2*a^2*b*p*x^2 + (a*b^2*p^2 + a*b^2*p)*x^4 + 2*a^3)*(b*x^2 + a)^p/(b^3*p
^3 + 6*b^3*p^2 + 11*b^3*p + 6*b^3)

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Sympy [A]  time = 5.32673, size = 979, normalized size = 13.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(b*x**2+a)**p,x)

[Out]

Piecewise((a**p*x**6/6, Eq(b, 0)), (2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5
*x**4) + 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + a**2/(4*a**2*b**3 +
8*a*b**4*x**2 + 4*b**5*x**4) + 4*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5
*x**4) + 4*a*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log
(-I*sqrt(a)*sqrt(1/b) + x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) + 2*b**2*x**4*log(I*sqrt(a)*sqrt(1/b) +
x)/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4) - 2*b**2*x**4/(4*a**2*b**3 + 8*a*b**4*x**2 + 4*b**5*x**4), Eq(
p, -3)), (-2*a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)
/(2*a*b**3 + 2*b**4*x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b*
*3 + 2*b**4*x**2) - 2*a*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2
*b**4*x**2), Eq(p, -2)), (a**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*b**3) + a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*b
**3) - a*x**2/(2*b**2) + x**4/(4*b), Eq(p, -1)), (2*a**3*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3
*p + 12*b**3) - 2*a**2*b*p*x**2*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p*
*2*x**4*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + a*b**2*p*x**4*(a + b*x**2)**p/(2*
b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + b**3*p**2*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 +
22*b**3*p + 12*b**3) + 3*b**3*p*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3) + 2*b
**3*x**6*(a + b*x**2)**p/(2*b**3*p**3 + 12*b**3*p**2 + 22*b**3*p + 12*b**3), True))

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Giac [B]  time = 1.47293, size = 312, normalized size = 4.33 \begin{align*} \frac{{\left (b x^{2} + a\right )}^{3}{\left (b x^{2} + a\right )}^{p} p^{2} - 2 \,{\left (b x^{2} + a\right )}^{2}{\left (b x^{2} + a\right )}^{p} a p^{2} +{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p} a^{2} p^{2} + 3 \,{\left (b x^{2} + a\right )}^{3}{\left (b x^{2} + a\right )}^{p} p - 8 \,{\left (b x^{2} + a\right )}^{2}{\left (b x^{2} + a\right )}^{p} a p + 5 \,{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p} a^{2} p + 2 \,{\left (b x^{2} + a\right )}^{3}{\left (b x^{2} + a\right )}^{p} - 6 \,{\left (b x^{2} + a\right )}^{2}{\left (b x^{2} + a\right )}^{p} a + 6 \,{\left (b x^{2} + a\right )}{\left (b x^{2} + a\right )}^{p} a^{2}}{2 \,{\left (b^{2} p^{3} + 6 \, b^{2} p^{2} + 11 \, b^{2} p + 6 \, b^{2}\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b*x^2+a)^p,x, algorithm="giac")

[Out]

1/2*((b*x^2 + a)^3*(b*x^2 + a)^p*p^2 - 2*(b*x^2 + a)^2*(b*x^2 + a)^p*a*p^2 + (b*x^2 + a)*(b*x^2 + a)^p*a^2*p^2
+ 3*(b*x^2 + a)^3*(b*x^2 + a)^p*p - 8*(b*x^2 + a)^2*(b*x^2 + a)^p*a*p + 5*(b*x^2 + a)*(b*x^2 + a)^p*a^2*p + 2
*(b*x^2 + a)^3*(b*x^2 + a)^p - 6*(b*x^2 + a)^2*(b*x^2 + a)^p*a + 6*(b*x^2 + a)*(b*x^2 + a)^p*a^2)/((b^2*p^3 +
6*b^2*p^2 + 11*b^2*p + 6*b^2)*b)